An absorption in an IR spectrum of an organic compound comes at 3100 cm^-1 and is assigned to an

step 1 Okay, so this question is about the vibration frequency. It asks about how does the vibration fr

An absorption in an IR spectrum of an organic compound comes...

Key Concepts

Harmonic Oscillator Model

The harmonic oscillator model is used to approximate the behavior of vibrating bonds in molecules. Within this model, the vibrational frequency is expressed as a function of the bond’s force constant and the reduced mass of the system. This theoretical framework explains why an increase in the reduced mass (as seen with deuteration) results in a lower vibrational frequency, thereby providing a quantitative basis for understanding experimental IR spectral shifts.

IR Spectroscopy

Infrared (IR) spectroscopy is a technique used to probe the vibrational transitions of molecules. It involves measuring the absorption of IR radiation, which causes bonds to stretch or bend. The frequency at which absorption occurs is characteristic of the types of bonds and their surrounding electronic environments, making it a useful tool for identifying functional groups in organic compounds.

Isotopic Substitution Effects

Isotopic substitution, such as replacing hydrogen with deuterium (deuteration), significantly affects vibrational frequencies. This happens because isotopes have different masses, altering the dynamics of the vibrating bond. Specifically, when a lighter isotope is replaced by a heavier one, the vibrational frequency typically decreases due to an increase in the reduced mass of the bond system.

Reduced Mass

Reduced mass is a critical concept in the study of molecular vibrations, defined as the product of the masses of two bonded atoms divided by their sum. In the context of bond vibrations, the vibrational frequency is inversely proportional to the square root of the reduced mass. When hydrogen is replaced by the heavier deuterium, the increased reduced mass leads to a decrease in the vibrational frequency, which is why shifts in IR spectra are observed upon deuteration.

Transcript

00:01 Okay, so this question is about the vibration frequency.

00:08 It asks about how does the vibration frequency change after you replace the hydrogen with deuterium, right? and there are actually three kinds of bond stretching, which is from c .h.

00:28 Stretching, right, to c .d.

00:31 Stretching.

00:33 And from the oh stretching to od stretching, and from nitrogen -hydrogen stretching to the nitrogen -duterium stretching.

00:51 So first, we need to know how to calculate vibration frequency for diametic molecules.

01:04 If you remember, you can recall in the chapter that about the ir spectroscopy, right, the vibration spectroscopy, we learned that if vibration, right, of a diametric molecule is actually equal to 1 over to pi times, right? k is a force constant over the effective mass, right? square root, right? constant mu is effective mass.

01:54 So from this, we can make some chance.

01:59 Suppose we have two vibrations, right, v1 over v2.

02:10 And we can just copy and paste over effective mass.

02:22 And for isotope replacement, usually we're seeing the isotope replacement, the fourth constant of bond, k, right, does not change.

02:37 So then, right, we can actually say, right, we can actually cancel 1 over 2 pi here.

02:48 And since the fourth constant is the same, we can also cancel them out.

02:52 And were actually equals to square root, right? the effect empty mass 2 over the affecting vast 1.

03:02 So you can say for isotope replacement, the ratio of the frequency is actually equals to the square root of the reverse ratio of their effective mass.

03:29 So now we can start our calculation.

03:35 So first, we can start to look at the frequency shift from c .h stretching to c .d.

03:49 So you can actually write down here that the frequency of the c .h.

03:58 Bond over the frequency of the cd bond.

04:03 Is actually equals to the square root of the effective mass of the cd molecule over the effective mass of the c .d molecule over the effective mass of the c .h molecule.

04:21 So the effective mass is actually equals to m1 times m2 over the atom 1 plus an atom 1 plus item 2.

04:38 So here we can actually write down here.

04:48 So the effective mass of the cd is actually equal to the mass of the carbon times the mass of the deuterium over the mass of the carbon plus the mass of the deuterium.

05:04 And this part has to over the mass of the carbon times the mass of the hydrogen over the mass of the carbon plus the mass of the hydrogen.

05:24 And here's the trick.

05:26 So normally when we calculate the frequency, we had to plug in the real mass of the atom.

05:33 But since you, if you can look at here, right, we'll actually calculate the ratio of the effect mass...

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